Let $\tau, \omega$ be two stopping times. How do we show that $\{\omega \ge \tau\} \in \mathscr{F}_{\tau -}$?
This result is used in the proof of Lemma 11 from : https://almostsuremath.com/2009/12/23/localization/
I am using the definition given here : https://almostsuremath.com/2009/11/23/sigma-algebras-at-a-stopping-time/
I need to show that $\{\omega \ge \tau\}$ can be generated by sets of the form $A \cap \{t < \tau\}$ for $t \ge 0$ and $A \in \mathscr{F}_t$ or by sets in $\mathscr{F}_0$. I think I need to use the rationals to come up with inequalities for $\omega$ and $\tau$ separately but I am stuck. I would greatly appreciate any help.
That is, $\mathscr{F}_{\tau -} = \sigma(\{A \cap \{t<\tau\}: t \ge 0, A \in \mathscr{F}_t\} \cup \mathscr{F}_0)$.
Updated: My solution. I think it is easier to show that $\{\omega < \tau \} \in \mathscr{F}_{\tau -}$. We have $\{\omega < \tau \} = \cup_{r \in \mathbb{Q}} \{\omega < r < \tau\}=\cup_r \{\{\omega < r\} \cap \{r<\tau\}\}$. And $\{ \omega < r \} \in \mathscr{F}_r$, so by definition above, we must have $\{\omega < \tau\} \in \mathscr{F}_{\tau -}$.
This appears to be false. If we let $\tau$ be any non-trivial stopping time taking strictly positive values, then setting $t = 0$ would give $$\{\tau \le \omega\} = \{\tau \le \omega\} \cap \{0 < \tau\} \in \mathcal F_0.$$ Since any constant is a stopping time, this implies $\{\tau \le c\} \in \mathcal F_0$ for all $c \in \mathbb{R}$, i.e. $\tau$ is $\mathcal F_0$ measurable. But we clearly have stopping times taking strictly positive values that aren't $\mathcal F_0$ measurable, e.g. the first hitting time of a Brownian motion.
Using the updated definition, we have that \begin{align*} \{\omega < \tau\} &= \bigcup_{t \in \mathbb{Q}}\left( \{\omega \le t\}\cap\{ t < \tau\}\right) \end{align*} and $\{\omega \le t\} \in \mathcal F_t$ because $\omega$ is a stopping time.